标签：ext 变量 ever next roo cto pop nbsp bfs
Given a binary tree, return the zigzag level order traversal of its nodes‘ values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
思路：还是层次遍历，只是增加一个int或bool类型变量标示当前行是否需要reverse。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector<queue<TreeNode*>> q(2); int curIndex = 0; int nextIndex = 1; int levelIndex = 0; vector<int> retItem; vector<vector<int>> ret; if(root) q[curIndex].push(root); while(!q[curIndex].empty()){ retItem.push_back(q[curIndex].front()->val); if(q[curIndex].front()->left) q[nextIndex].push(q[curIndex].front()->left); if(q[curIndex].front()->right) q[nextIndex].push(q[curIndex].front()->right); q[curIndex].pop(); if(q[curIndex].empty()){ //end of this level if(levelIndex) reverse(retItem.begin(), retItem.end()); ret.push_back(retItem); retItem.clear(); curIndex = (curIndex+1) & 0x01; nextIndex = (nextIndex+1) & 0x01; levelIndex = (levelIndex+1) & 0x01; } } return ret; } };
103. Binary Tree Zigzag Level Order Traversal (Tree, Queue; BFS)
标签：ext 变量 ever next roo cto pop nbsp bfs
原文地址：http://www.cnblogs.com/qionglouyuyu/p/6067497.html
